What is the sum from n=1 to infinity of-3(8/7)^{n-1} ?

The sum from n=1 to infinity of-3(8/7)^{n-1} is diverges

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sum from n=1 to infinity of-3(8/7)^{n-1}

n = 1 \sum \infty - 3 (\frac{8}{7})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty - 3 (\frac{8}{7})^{n - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - 3 \cdot n = 1 \sum \infty (\frac{8}{7})^{n - 1}

Apply Series Ratio Test:

diverges

= - 3 diverges

= diverges

n = 0 \sum \infty \frac{1 + ( - 1 ) ^{n}}{n}

n = 0 \sum \infty 8^{\frac{1}{n}}

n = 1 \sum \infty \frac{4}{e ^{n}}

n = 1 \sum \infty \frac{1}{( 7 n + 4 ) ^{2}}

n = 0 \sum \infty n^{5}

What is the sum from n=1 to infinity of-3(8/7)^{n-1} ?
The sum from n=1 to infinity of-3(8/7)^{n-1} is diverges